In a class of $8$, there are $5$ students who will be eaten by bears. If the teacher chooses $3$ students, what is the probability that none of the three of them will be eaten by bears?
We can think about this problem as the probability of $3$ events happening. The first event is the teacher choosing one student who will not be eaten by bears. The second event is the teacher choosing another student who will not be eaten by bears, given that the teacher already chose someone who will not be eaten by bears, and so on. The probabilty that the teacher will choose someone who will not be eaten by bears is the number of students who will not be eaten by bears divided by the total number of students: $\dfrac{3} {8}$ Once the teacher's chosen one student, there are only $7$ left. There's also one fewer student who will not be eaten by bears, since the teacher isn't going to pick the same student twice. So, the probability that the teacher picks a second student who also will not be eaten by bears is $\dfrac{2} {7}$ The probability of the teacher picking two students who will not be eaten by bears must then be $\dfrac{3} {8} \cdot \dfrac{2} {7}$ We can continue using the same logic for the rest of the students the teacher picks. So, the probability of the teacher picking $3$ students such that none of them will be eaten by bears is $\dfrac{3}{8}\cdot\dfrac{2}{7}\cdot\dfrac{1}{6} = \dfrac{1}{56}$